Regular pentagon from a square paper

This doesn't produce a perfect pentagon. In the right triangle at the lower right corner of the rectangle, call the smaller angle $\alpha$ and the larger angle $\beta$. When you valley fold at $\heartsuit-\diamondsuit$ so that the stars coincide, this creates the angles indicated below.

Here's how it looks after the valley fold:

The next step (the mountain fold) hides the angle $\beta-\alpha$ behind the rest of the paper. The valley fold that follows will bisect the angle $\alpha + 45^\circ$. So when the origami is complete the folds you've made will create five angles in a half circle: one with measure $\beta-\alpha$ and four with measure $\frac12(\alpha + 45^\circ)$.

In a perfect pentagon, these angles would all be equal, i.e., $$ \beta - \alpha = \textstyle\frac12(\alpha + 45^\circ). $$ Solving this, knowing that $\alpha + \beta=90^\circ$, this means in a regular pentagon we must have $\alpha=27^\circ$ and $\beta=63^\circ$. In particular this requires $$ \tan\alpha=\tan(27^\circ) \approx 0.5095.$$ But for the advertised construction, we have $\tan\alpha = 0.5$. Close, but not exact!